Examples On Integration By Substitution Set-4

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Example – 18

Evaluate the following integrals:

(a) \(\begin{align} \int {\frac{1}{{\sqrt {{a^2} - {x^2}} }}dx} J \end{align}\) (b) \(\begin{align} \int {\frac{1}{{{a^2} + {x^2}}}dx}\end{align}\) (c) \(\begin{align} \int {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}dx} \end{align}\)
(d) \(\begin{align} \int {\frac{1}{{{x^2} - {a^2}}}dx} \end{align}\) (e) \(\begin{align}\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx} \end{align}\) (f) \(\begin{align}\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx}  \end{align}\)

Solution: This example is very important in the sense that the techniques subsequently described to evaluate these integrals can be used anywhere where such expressions are encountered.

Recall that the results to parts-(a), (b) and (c) have already been mentioned in the table titled ‘Basic integration formulae’ on page -2. Also we have already seen (in examples 12, 13, 15), some of the integrals of these forms. Before starting with the solutions, consider the following table carefully which describes certain substitutions that can be used whenever expressions of the forms above are encountered.

  \[{{\rm{expression}}}\] \[{{\rm{can\; be\; reduced\; by\; the\; substitution}}}\]
\[{(1)}\] \[{(2)}\] \[{(3)}\] \[{{a^2} - {x^2}}\] \[{{a^2} + {x^2}}\] \[{{x^2} - {a^2}}\] \[{x = a\sin \theta \,\,{\rm{or}}\,\,x = a\cos \theta }\] \[{x = a\tan \theta \,\,{\rm{or}}\,\,x = a\cot \theta }\] \[{\,x = a\sec \theta \,\,{\rm{or}}\,\,x = a\,{\rm{cosec}}\theta }\]

Verify that these substitutions will reduce the corresponding algebraic expressions to simpler trigonometric expressions. We now proceed with the solutions:

(a) \(\begin{align} I = \int {\frac{1}{{\sqrt {{a^2} - {x^2}} }} dx} \end{align}\)

\[\begin{align}& \text{Substitute}\ \ \ \ x=a\sin \theta  \\ & \qquad\quad\Rightarrow \ \ dx=a\cos \theta \;d\theta  \\ & and\ \ \;\text{ }{{a}^{2}}-{{x}^{2}}\;={{a}^{2}}-{{a}^{2}}{{\sin }^{2}}\theta \text{ } \\ & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; ={{a}^{2}}{{\cos }^{2}}\theta  \\ &\quad \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ I=\int{\frac{a\cos \theta }{a\cos \theta }d\theta } \\  & \qquad\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,=\int{d\theta } \\ &\qquad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\theta +C \\ & \qquad\qquad\qquad={{\sin }^{-1}}\frac{x}{a}+C\\\end{align}\]

Notice how convenient the integral became with the mentioned substitution.

(b) \(\begin{align} I = \int {\frac{1}{{{a^2} + {x^2}}}dx}\end{align}\)

\[\begin{align}& \text{Substitute}\;x=a\tan \theta  \\ &\quad\;\;\, \Rightarrow\quad dx=a\;{{\sec }^{2}}\theta\; d\theta  \\ & \text{and}\;\;{{a}^{2}}+{{x}^{2}}={{a}^{2}}+{{a}^{2}}{{\tan }^{2}}\theta  \\  & \qquad\qquad\quad\;={{a}^{2}}{{\sec }^{2}}\theta  \\ & \quad\quad\Rightarrow\quad I=\int{\frac{a{{\sec }^{2}}\theta }{{{a}^{2}}{{\sec }^{2}}\theta }d\theta } \\ &\qquad\qquad\quad =\frac{1}{a}d\theta  \\ & \qquad\qquad\quad=\frac{\theta }{a}+C \\ &\qquad\qquad\quad =\frac{1}{a}{{\tan }^{-1}}\frac{x}{a}+C \\ \end{align}\]

(c) \(\begin{align}I = \int {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}} dx\end{align}\)

\[\begin{align}& \text{Substitute}\;\;\;\;x=a\sec \theta  \\  &\;\; \qquad\Rightarrow \quad dx=a\sec \theta \tan \theta d\theta  \\  & \,And \quad{{x}^{2}}-{{a}^{2}}={{a}^{2}}{{\sec }^{2}}\theta -{{a}^{2}} \\&\qquad\qquad\qquad\; ={{a}^{2}}{{\tan }^{2}}\theta  \\ &\qquad\quad \Rightarrow\quad I=\int{\frac{a\sec \theta \tan \theta }{a\sec \theta \sqrt{{{a}^{2}}{{\tan }^{2}}\theta }}d\theta } \\ & \qquad\qquad\qquad=\frac{1}{a}\int{d\theta } \\ &\qquad\qquad\qquad =\frac{\theta }{a}+C \\ & \qquad\qquad\qquad=\frac{1}{a}{{\sec }^{-1}}\frac{x}{a}+C \\ \end{align}\]

(d) \(\begin{align}I = \int {\frac{1}{{{x^2} - {a^2}}}dx}\end{align}\)

\[\begin{align}& \text{Substitute} \qquad x=a\sec \theta  \\ & \qquad \quad\;\Rightarrow\quad dx=a\sec \theta \tan \theta \;d\theta  \\ & \quad and \;\;\;\;{{x}^{2}}-{{a}^{2}}={{a}^{2}}{{\tan }^{2}}\theta  \\ & \qquad\qquad\Rightarrow \quad I=\int{\frac{a\sec \theta \tan \theta }{{{a}^{2}}{{\tan }^{2}}\theta }}d\theta  \\  &\qquad\qquad\qquad\quad =\frac{1}{a}\int{\text{cosec}\theta \;d\theta } \\ &\qquad\qquad\qquad\quad =\frac{1}{a}\ln \left| \text{cosec}\theta -\cot \theta  \right|+C \\  &\qquad\qquad\qquad\quad  =\frac{1}{a}\ln \left| \frac{x}{\sqrt{{{x}^{2}}-{{a}^{2}}}}-\frac{a}{\sqrt{{{x}^{2}}-{{a}^{2}}}} \right|+C \\  &\qquad\qquad\qquad\quad  =\frac{1}{a}\ln \left| \sqrt{\frac{x-a}{x+a}} \right|+C \\  &\qquad\qquad\qquad\quad  =\frac{1}{2a}\ln \left| \frac{x-a}{x+a} \right|+C \\ \end{align}\]

Instead of the substitution technique as described above, this integral could have alternatively been evaluate much more simply by a straightforward rearrangement of the expression:

\[\begin{align}&\;I = \int {\frac{1}{{{x^2} - {a^2}}}dx}\\&\,\,\,\, = \frac{1}{{2a}}\int {\left( {\frac{1}{{x - a}} - \frac{1}{{x + a}}} \right)dx} \\&\,\,\,\, = \frac{1}{{2a}}\left( {\ln (x - a) - \ln (x + a)} \right) + C\\ &\,\,\,\, = \frac{1}{{2a}}\ln \left( {\frac{{x - a}}{{x + a}}} \right) + C\end{align}\]

This again shows that there is no set rule to integration. You have to use your intuition to figure out the shortest possible route to the final answer.

(e) \(\begin{align}I = \int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx} \end{align}\)

\[\begin{align}& Substitute\,\; x=a\tan \theta  \\ &\qquad \Rightarrow\quad dx=a{{\sec }^{2}}\theta d\theta  \\ & and\quad{{x}^{2}}+{{a}^{2}}={{a}^{2}}{{\tan }^{2}}\theta +{{a}^{2}} \\ &\qquad \qquad\quad\;\;\;={{a}^{2}}{{\sec }^{2}}\theta  \\ &\quad\quad \;\;\;\Rightarrow\quad I=\int{\frac{a{{\sec }^{2}}\theta }{\sqrt{{{a}^{2}}{{\sec }^{2}}\theta }}d\theta } \\  &\qquad \qquad\quad\;\;\;=\int{\sec \theta d\theta } \\ &\qquad \qquad\quad\;\;\; =\ln \left| \sec \theta +\tan \theta  \right|+C \\ & \qquad \qquad\quad\;\;\;=\ln \left| \sqrt{1+{{\tan }^{2}}\theta }+\tan \theta  \right|+C \\ &\qquad \qquad\quad\;\;\; =\ln \left| \frac{\sqrt{{{x}^{2}}+{{a}^{2}}}}{a}+\frac{x}{a} \right|+C \\ &\qquad \qquad\quad\;\;\; =\ln \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|-\ln a+C \\  &\qquad \qquad\quad\;\;\; =\ln \left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|+{C}' \\ \end{align}\]

(f)\(\begin{align} I = \int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx}  \end{align}\)

\[\begin{align} & \text{Substitute} \;\;x=a\sec \theta \\&\qquad \Rightarrow \quad dx = a\sec \theta \tan \theta d\theta \\ &and\quad {x^2} - {a^2} = {a^2}{\sec ^2}\theta - {a^2}\\ &\qquad\qquad\quad\;\;= {a^2}{\tan ^2}\theta\\ &\quad\quad\;\; \Rightarrow\quad I=\int{\frac{a\sec \theta \tan \theta }{\sqrt{{{a}^{2}}{{\tan }^{2}}\theta }}d\theta } \\ &\qquad\qquad \quad\;=\int{\sec \theta d\theta } \\ &\qquad\qquad \quad\;=\ln \left| \sec \theta +\tan \theta  \right|+C \\  &\qquad\qquad \quad\; =\ln \left| \sec \theta +\sqrt{{{\sec }^{2}}\theta -1} \right|+C \\ &\qquad\qquad \quad\; =\ln \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+{C}' \\ \end{align}\]

The expressions encountered in these six examples will widely be found elsewhere in this chapter too and therefore, you are advised to commit these six results to memory. If memorization is not possible for you, you should at least understand the techniques involved carefully so that you are quickly able to reproduce the answers whenever the need arises.

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Download SOLVED Practice Questions of Examples On Integration By Substitution Set-4 for FREE
Indefinite Integration
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